Series: determining convergence of series problem












1












$begingroup$


I am working on the following assignment about series:
$$ a_n , , is ,, a ,, convergent ,, progression ,, and \
sum_1^infty (a_n+1) = lim_{xtoinfty} {x^2-x+1over x^2-cosx}$$ Determine if the sum converges and $$lim_{xtoinfty} a_n$$



What I have done so far:
I know that for a sum to converge, its associated progression must have 0 as its limit at infinity (the neccessary condition).
I also know that, if the limit of the sum at infinity exists, then the sum converges and that limit is the value of the sum. My first step was finding the RHS limit:
$$lim_{xtoinfty} {x^2-x+1over x^2-cosx} = 1 $$

Now, this means -to me- that the sum is convergent, and it converges to 1 . However, I find the following problem. I also know that summations are "linear", I mean:
$$sum_1^infty a_n+1 = sum_1^infty a_n+ sum_1^infty 1 $$



I know the first one is convergent, but the second one clearly is not! Its limit at infinity is infinity; and anything + infinity = infinity = divergent. I cannot decide where was my mistake. Could someone help me to figure it out?



Lots of thanks in advance!






sequences-and-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    See my answer below.
    $endgroup$
    – John Omielan
    Dec 27 '18 at 1:54








  • 1




    $begingroup$
    Perhaps the progression ${a_n}$ converges to $-1$ : $$limlimits_{ntoinfty}a_n=-1$$
    $endgroup$
    – rsadhvika
    Dec 27 '18 at 1:59










  • $begingroup$
    $$sum_1^infty( -1+1)~~~ ?=~~~ sum_1^infty -1+ sum_1^infty 1$$
    $endgroup$
    – rsadhvika
    Dec 27 '18 at 2:00












  • $begingroup$
    @rsadhvika I initially thought the same thing but, wouldn't that contradict the first limit found?
    $endgroup$
    – Artem
    Dec 27 '18 at 2:21








  • 1




    $begingroup$
    You knew $$sum_1^infty (a_n+1 )= 1$$
    $endgroup$
    – rsadhvika
    Dec 27 '18 at 2:30
















1












$begingroup$


I am working on the following assignment about series:
$$ a_n , , is ,, a ,, convergent ,, progression ,, and \
sum_1^infty (a_n+1) = lim_{xtoinfty} {x^2-x+1over x^2-cosx}$$ Determine if the sum converges and $$lim_{xtoinfty} a_n$$



What I have done so far:
I know that for a sum to converge, its associated progression must have 0 as its limit at infinity (the neccessary condition).
I also know that, if the limit of the sum at infinity exists, then the sum converges and that limit is the value of the sum. My first step was finding the RHS limit:
$$lim_{xtoinfty} {x^2-x+1over x^2-cosx} = 1 $$

Now, this means -to me- that the sum is convergent, and it converges to 1 . However, I find the following problem. I also know that summations are "linear", I mean:
$$sum_1^infty a_n+1 = sum_1^infty a_n+ sum_1^infty 1 $$



I know the first one is convergent, but the second one clearly is not! Its limit at infinity is infinity; and anything + infinity = infinity = divergent. I cannot decide where was my mistake. Could someone help me to figure it out?



Lots of thanks in advance!






sequences-and-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    See my answer below.
    $endgroup$
    – John Omielan
    Dec 27 '18 at 1:54








  • 1




    $begingroup$
    Perhaps the progression ${a_n}$ converges to $-1$ : $$limlimits_{ntoinfty}a_n=-1$$
    $endgroup$
    – rsadhvika
    Dec 27 '18 at 1:59










  • $begingroup$
    $$sum_1^infty( -1+1)~~~ ?=~~~ sum_1^infty -1+ sum_1^infty 1$$
    $endgroup$
    – rsadhvika
    Dec 27 '18 at 2:00












  • $begingroup$
    @rsadhvika I initially thought the same thing but, wouldn't that contradict the first limit found?
    $endgroup$
    – Artem
    Dec 27 '18 at 2:21








  • 1




    $begingroup$
    You knew $$sum_1^infty (a_n+1 )= 1$$
    $endgroup$
    – rsadhvika
    Dec 27 '18 at 2:30














1












1








1





$begingroup$


I am working on the following assignment about series:
$$ a_n , , is ,, a ,, convergent ,, progression ,, and \
sum_1^infty (a_n+1) = lim_{xtoinfty} {x^2-x+1over x^2-cosx}$$ Determine if the sum converges and $$lim_{xtoinfty} a_n$$



What I have done so far:
I know that for a sum to converge, its associated progression must have 0 as its limit at infinity (the neccessary condition).
I also know that, if the limit of the sum at infinity exists, then the sum converges and that limit is the value of the sum. My first step was finding the RHS limit:
$$lim_{xtoinfty} {x^2-x+1over x^2-cosx} = 1 $$

Now, this means -to me- that the sum is convergent, and it converges to 1 . However, I find the following problem. I also know that summations are "linear", I mean:
$$sum_1^infty a_n+1 = sum_1^infty a_n+ sum_1^infty 1 $$



I know the first one is convergent, but the second one clearly is not! Its limit at infinity is infinity; and anything + infinity = infinity = divergent. I cannot decide where was my mistake. Could someone help me to figure it out?



Lots of thanks in advance!






sequences-and-series







share|cite|improve this question











$endgroup$




I am working on the following assignment about series:
$$ a_n , , is ,, a ,, convergent ,, progression ,, and \
sum_1^infty (a_n+1) = lim_{xtoinfty} {x^2-x+1over x^2-cosx}$$ Determine if the sum converges and $$lim_{xtoinfty} a_n$$



What I have done so far:
I know that for a sum to converge, its associated progression must have 0 as its limit at infinity (the neccessary condition).
I also know that, if the limit of the sum at infinity exists, then the sum converges and that limit is the value of the sum. My first step was finding the RHS limit:
$$lim_{xtoinfty} {x^2-x+1over x^2-cosx} = 1 $$

Now, this means -to me- that the sum is convergent, and it converges to 1 . However, I find the following problem. I also know that summations are "linear", I mean:
$$sum_1^infty a_n+1 = sum_1^infty a_n+ sum_1^infty 1 $$



I know the first one is convergent, but the second one clearly is not! Its limit at infinity is infinity; and anything + infinity = infinity = divergent. I cannot decide where was my mistake. Could someone help me to figure it out?



Lots of thanks in advance!






sequences-and-series




sequences-and-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 27 '18 at 2:23







Artem

















asked Dec 27 '18 at 1:49









ArtemArtem

425




425












  • $begingroup$
    See my answer below.
    $endgroup$
    – John Omielan
    Dec 27 '18 at 1:54








  • 1




    $begingroup$
    Perhaps the progression ${a_n}$ converges to $-1$ : $$limlimits_{ntoinfty}a_n=-1$$
    $endgroup$
    – rsadhvika
    Dec 27 '18 at 1:59










  • $begingroup$
    $$sum_1^infty( -1+1)~~~ ?=~~~ sum_1^infty -1+ sum_1^infty 1$$
    $endgroup$
    – rsadhvika
    Dec 27 '18 at 2:00












  • $begingroup$
    @rsadhvika I initially thought the same thing but, wouldn't that contradict the first limit found?
    $endgroup$
    – Artem
    Dec 27 '18 at 2:21








  • 1




    $begingroup$
    You knew $$sum_1^infty (a_n+1 )= 1$$
    $endgroup$
    – rsadhvika
    Dec 27 '18 at 2:30


















  • $begingroup$
    See my answer below.
    $endgroup$
    – John Omielan
    Dec 27 '18 at 1:54








  • 1




    $begingroup$
    Perhaps the progression ${a_n}$ converges to $-1$ : $$limlimits_{ntoinfty}a_n=-1$$
    $endgroup$
    – rsadhvika
    Dec 27 '18 at 1:59










  • $begingroup$
    $$sum_1^infty( -1+1)~~~ ?=~~~ sum_1^infty -1+ sum_1^infty 1$$
    $endgroup$
    – rsadhvika
    Dec 27 '18 at 2:00












  • $begingroup$
    @rsadhvika I initially thought the same thing but, wouldn't that contradict the first limit found?
    $endgroup$
    – Artem
    Dec 27 '18 at 2:21








  • 1




    $begingroup$
    You knew $$sum_1^infty (a_n+1 )= 1$$
    $endgroup$
    – rsadhvika
    Dec 27 '18 at 2:30
















$begingroup$
See my answer below.
$endgroup$
– John Omielan
Dec 27 '18 at 1:54






$begingroup$
See my answer below.
$endgroup$
– John Omielan
Dec 27 '18 at 1:54






1




1




$begingroup$
Perhaps the progression ${a_n}$ converges to $-1$ : $$limlimits_{ntoinfty}a_n=-1$$
$endgroup$
– rsadhvika
Dec 27 '18 at 1:59




$begingroup$
Perhaps the progression ${a_n}$ converges to $-1$ : $$limlimits_{ntoinfty}a_n=-1$$
$endgroup$
– rsadhvika
Dec 27 '18 at 1:59












$begingroup$
$$sum_1^infty( -1+1)~~~ ?=~~~ sum_1^infty -1+ sum_1^infty 1$$
$endgroup$
– rsadhvika
Dec 27 '18 at 2:00






$begingroup$
$$sum_1^infty( -1+1)~~~ ?=~~~ sum_1^infty -1+ sum_1^infty 1$$
$endgroup$
– rsadhvika
Dec 27 '18 at 2:00














$begingroup$
@rsadhvika I initially thought the same thing but, wouldn't that contradict the first limit found?
$endgroup$
– Artem
Dec 27 '18 at 2:21






$begingroup$
@rsadhvika I initially thought the same thing but, wouldn't that contradict the first limit found?
$endgroup$
– Artem
Dec 27 '18 at 2:21






1




1




$begingroup$
You knew $$sum_1^infty (a_n+1 )= 1$$
$endgroup$
– rsadhvika
Dec 27 '18 at 2:30




$begingroup$
You knew $$sum_1^infty (a_n+1 )= 1$$
$endgroup$
– rsadhvika
Dec 27 '18 at 2:30










1 Answer
1






active

oldest

votes


















1












$begingroup$

I believe the expression



$$sum_{1}^{infty} a_n + 1$$



means



$$left(sum_{1}^{infty} a_n right) + 1$$



If it meant as you suggest, the expression would instead most likely be written as



$$sum_{1}^{infty} left( a_n + 1 right)$$



This would avoid potential ambiguity.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Hey, thanks for the answer. Nope, I asked the lecturer and he said it must be considered with brackets!
    $endgroup$
    – Artem
    Dec 27 '18 at 2:10










  • $begingroup$
    @Artem You are welcome. You may wish to suggest to your lecturer that he/she uses brackets to avoid this potential ambiguity.
    $endgroup$
    – John Omielan
    Dec 27 '18 at 2:13










  • $begingroup$
    @Artem Well, the sum converges if it is equal to 1...
    $endgroup$
    – Osvaldo Paniccia
    Dec 27 '18 at 2:27











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

I believe the expression



$$sum_{1}^{infty} a_n + 1$$



means



$$left(sum_{1}^{infty} a_n right) + 1$$



If it meant as you suggest, the expression would instead most likely be written as



$$sum_{1}^{infty} left( a_n + 1 right)$$



This would avoid potential ambiguity.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Hey, thanks for the answer. Nope, I asked the lecturer and he said it must be considered with brackets!
    $endgroup$
    – Artem
    Dec 27 '18 at 2:10










  • $begingroup$
    @Artem You are welcome. You may wish to suggest to your lecturer that he/she uses brackets to avoid this potential ambiguity.
    $endgroup$
    – John Omielan
    Dec 27 '18 at 2:13










  • $begingroup$
    @Artem Well, the sum converges if it is equal to 1...
    $endgroup$
    – Osvaldo Paniccia
    Dec 27 '18 at 2:27
















1












$begingroup$

I believe the expression



$$sum_{1}^{infty} a_n + 1$$



means



$$left(sum_{1}^{infty} a_n right) + 1$$



If it meant as you suggest, the expression would instead most likely be written as



$$sum_{1}^{infty} left( a_n + 1 right)$$



This would avoid potential ambiguity.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Hey, thanks for the answer. Nope, I asked the lecturer and he said it must be considered with brackets!
    $endgroup$
    – Artem
    Dec 27 '18 at 2:10










  • $begingroup$
    @Artem You are welcome. You may wish to suggest to your lecturer that he/she uses brackets to avoid this potential ambiguity.
    $endgroup$
    – John Omielan
    Dec 27 '18 at 2:13










  • $begingroup$
    @Artem Well, the sum converges if it is equal to 1...
    $endgroup$
    – Osvaldo Paniccia
    Dec 27 '18 at 2:27














1












1








1





$begingroup$

I believe the expression



$$sum_{1}^{infty} a_n + 1$$



means



$$left(sum_{1}^{infty} a_n right) + 1$$



If it meant as you suggest, the expression would instead most likely be written as



$$sum_{1}^{infty} left( a_n + 1 right)$$



This would avoid potential ambiguity.






share|cite|improve this answer











$endgroup$



I believe the expression



$$sum_{1}^{infty} a_n + 1$$



means



$$left(sum_{1}^{infty} a_n right) + 1$$



If it meant as you suggest, the expression would instead most likely be written as



$$sum_{1}^{infty} left( a_n + 1 right)$$



This would avoid potential ambiguity.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 27 '18 at 2:05

























answered Dec 27 '18 at 1:58









John OmielanJohn Omielan

3,9251215




3,9251215












  • $begingroup$
    Hey, thanks for the answer. Nope, I asked the lecturer and he said it must be considered with brackets!
    $endgroup$
    – Artem
    Dec 27 '18 at 2:10










  • $begingroup$
    @Artem You are welcome. You may wish to suggest to your lecturer that he/she uses brackets to avoid this potential ambiguity.
    $endgroup$
    – John Omielan
    Dec 27 '18 at 2:13










  • $begingroup$
    @Artem Well, the sum converges if it is equal to 1...
    $endgroup$
    – Osvaldo Paniccia
    Dec 27 '18 at 2:27


















  • $begingroup$
    Hey, thanks for the answer. Nope, I asked the lecturer and he said it must be considered with brackets!
    $endgroup$
    – Artem
    Dec 27 '18 at 2:10










  • $begingroup$
    @Artem You are welcome. You may wish to suggest to your lecturer that he/she uses brackets to avoid this potential ambiguity.
    $endgroup$
    – John Omielan
    Dec 27 '18 at 2:13










  • $begingroup$
    @Artem Well, the sum converges if it is equal to 1...
    $endgroup$
    – Osvaldo Paniccia
    Dec 27 '18 at 2:27
















$begingroup$
Hey, thanks for the answer. Nope, I asked the lecturer and he said it must be considered with brackets!
$endgroup$
– Artem
Dec 27 '18 at 2:10




$begingroup$
Hey, thanks for the answer. Nope, I asked the lecturer and he said it must be considered with brackets!
$endgroup$
– Artem
Dec 27 '18 at 2:10












$begingroup$
@Artem You are welcome. You may wish to suggest to your lecturer that he/she uses brackets to avoid this potential ambiguity.
$endgroup$
– John Omielan
Dec 27 '18 at 2:13




$begingroup$
@Artem You are welcome. You may wish to suggest to your lecturer that he/she uses brackets to avoid this potential ambiguity.
$endgroup$
– John Omielan
Dec 27 '18 at 2:13












$begingroup$
@Artem Well, the sum converges if it is equal to 1...
$endgroup$
– Osvaldo Paniccia
Dec 27 '18 at 2:27




$begingroup$
@Artem Well, the sum converges if it is equal to 1...
$endgroup$
– Osvaldo Paniccia
Dec 27 '18 at 2:27


















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